Question 174: Properties of Matter

Question

When an air bubble of radius r rises from the bottom to the surface of a lake its radius becomes $${{5r} \over 4}.$$ Taking the atmospheric pressure to be equal to 10 m height of water column, the depth of the lake would approximately be (ignore the surface tension and the effect of temperature) :

Accepted Answer

Given, bubble of radius r rises from bottom of lake. The radius of bubble at top becomes 5r/4. Therefore, pressure in bubble at bottom is $${P_1} = {P_0} + ho gh + {{4T} \over r}$$
Pressure in bubble at top is $${P_2} = {P_0} + {{4T} \over {5r/4}}$$
Now, we know $${P_1}{V_1} = {P_2}{V_2}$$, therefore,
$$\left( {{P_0} + ho gh + {{4T} \over 4}} ight){{4\pi } \over 3}{r^3} = \left( {{P_0} + {{4T} \over {5r/4}}} ight){{4\pi } \over 3}{\left( {{{5r} \over 4}} ight)^3}$$
$${P_0} + ho hg + {{4T} \over r} = \left( {{P_0} + {{4T 	imes 4} \over {5r}}} ight){{125} \over {64}}$$
Now given P
0
= 10$$ho$$g, therefore,
$$10ho g + ho gh + {{4T} \over r} = {{125} \over {64}} 	imes 10ho g + {{16T} \over {5r}} 	imes {{125} \over {64}}$$
Neglecting effect of temperature, we get
$$10ho g + ho gh = {{125} \over {64}} 	imes 10ho g$$
$$ \Rightarrow 10 + h = {{125} \over {64}} 	imes 10 \Rightarrow h = {{1250} \over {64}} - 10 = 9.53125\,m$$
$$\Rightarrow$$ h $$\sim$$ 9.5 m

Answer

11.2 m

Answer

8.7 m

Answer

9.5 m

Answer

10.5 m

Properties of Matter

When an air bubble of radius r rises from the bottom to the surface of a lake its radius becomes ${{5r} \over 4}.$ Taking the atmospheric pressure to be equal to 10 m height of water column, the depth of the lake would approximately be (ignore the surface tension and the effect of temperature) :

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